F per unit length = μ₀ I₁ I₂ / (2π d). Same direction → attractive; opposite → repulsive. Defines ampere: F = 2 × 10⁻⁷ N/m for I₁ = I₂ = 1 A, d = 1 m.
-- NCERT Class 12 Physics, Ch. 4, p. 123Force Between Parallel Currents
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 24 Sep 2026
Force Between Parallel Currents, explained for NEET
The sign convention is where marks go. Two long parallel wires carrying current in the same direction attract; in opposite directions they repel. This is the reverse of the electrostatic habit — like charges repel, but like currents attract — and under time pressure that habit overwrites the correct rule.
The mechanism, not the mnemonic: wire 1 sets up a field at wire 2's location. Wire 2 sits in that field and feels a force given by the current-in-a-field rule. Combining the two gives the standard result in NCERT Class 12 Physics, Chapter 4, page 123:
F/L = μ₀ I₁ I₂ / (2πd)
Three features of this expression decide most questions.
First, it is a force per unit length, not a force. The wires are treated as long (effectively infinite), so the total force grows with however much length you consider. A question asking for force in newtons must supply a length; if it does not, the answer is in N/m.
Second, both currents appear to the first power, and the separation to the first power in the denominator — not squared. Doubling one current doubles the force. Doubling the separation halves it. An inverse-square guess, imported from Coulomb, is a standard wrong option.
Third, the force on the two wires is equal in magnitude and opposite in direction, whatever the currents are. Even when I₁ = 10 A and I₂ = 1 A, each wire feels the same F/L, because the product I₁I₂ is symmetric. Newton's third law is not suspended by the current being unequal.
Bridge to NEET: this topic appears roughly once every two or three papers, usually as a direct substitution or a proportionality comparison, and it is the physical basis of the SI ampere (covered separately in this unit).
Watch out for: reading "opposite directions" and still writing attraction, and for stems that give a length so the answer is a force rather than a force per unit length.
Can you answer these Force Between Parallel Currents MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Two long straight parallel wires carry steady currents in the same direction. The force between them is
Show answer and why every option is right or wrong
Answer: A. Parallel currents in the same direction attract; antiparallel currents repel. This is stated with the force expression in NCERT Class 12 Physics, Chapter 4, page 123.
Why B is wrong: B is wrong because repulsion occurs for currents in opposite directions, not the same direction; this is the electrostatic 'like repels like' habit carried across incorrectly.
Why C is wrong: C is wrong because the force vanishes only if one of the currents is zero or the separation is infinite; two finite steady currents always interact.
Why D is wrong: D is wrong because the attraction depends on the sense of the currents, not on their equality; F/L = μ₀I₁I₂/(2πd) is attractive for any same-direction pair.
MCQ 2Easy RecallPractice
In the expression F/L = μ₀I₁I₂/(2πd) for two long parallel conductors, the quantity on the left-hand side has SI units of
Show answer and why every option is right or wrong
Answer: B. The expression gives force per unit length, so its SI unit is N/m. The per-unit-length form is used because the wires are idealised as infinitely long (NCERT Class 12 Physics, Chapter 4, page 123).
Why A is wrong: A is wrong because a plain newton would be the total force, which requires multiplying F/L by a specified length of wire; the formula as written does not give that.
Why C is wrong: C is wrong because N·m is the unit of torque or work, obtained by multiplying force by a distance rather than dividing by one.
Why D is wrong: D is wrong because the tesla is the unit of magnetic field B, not of the mechanical force per length between the wires.
MCQ 3Direct ApplicationPractice
Two long straight parallel wires in vacuum are separated by 0.100 m. They carry steady currents of 4.00 A and 6.00 A in opposite directions. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the magnitude of the force per unit length between them is
Show answer and why every option is right or wrong
Answer: A. F/L = (4π × 10⁻⁷ × 4.00 × 6.00)/(2π × 0.100) = (2 × 10⁻⁷ × 24.0)/0.100 = 4.80 × 10⁻⁵ N/m; opposite currents repel (NCERT Class 12 Physics, Chapter 4, page 123).
Why B is wrong: B is wrong on direction only: the arithmetic is right, but currents in opposite directions repel. Same-direction currents are the attracting case.
Why C is wrong: C is wrong because it halves the correct product — a slip of using μ₀/(4πd) instead of μ₀/(2πd), i.e. dropping a factor of 2 from the numerator.
Why D is wrong: D is wrong by one power of ten, the signature of substituting d = 0.0100 m instead of the stated 0.100 m.
MCQ 4Direct ApplicationPractice
Two long parallel wires carry currents I₁ and I₂ and experience a force per unit length F₀. The separation is now doubled while both currents are unchanged. The new force per unit length is
Show answer and why every option is right or wrong
Answer: D. F/L ∝ 1/d, so doubling d halves the force per unit length to F₀/2. The separation enters to the first power, not squared (NCERT Class 12 Physics, Chapter 4, page 123).
Why A is wrong: A is wrong because it treats the force as increasing with separation, which reverses the inverse dependence entirely.
Why B is wrong: B is wrong for the same reason — the force falls when the wires are moved apart, it does not grow.
Why C is wrong: C is wrong because F₀/4 assumes an inverse-square law imported from Coulomb's law; the parallel-wire force goes as 1/d.
MCQ 5Direct ApplicationPractice
Two long parallel wires 0.200 m apart carry currents of 10.0 A and 1.00 A in the same direction. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the force per unit length on the 1.00 A wire is
Show answer and why every option is right or wrong
Answer: C. F/L = (2 × 10⁻⁷ × 10.0 × 1.00)/0.200 = 1.00 × 10⁻⁵ N/m, and the expression is symmetric in I₁ and I₂, so both wires feel the same magnitude (NCERT Class 12 Physics, Chapter 4, page 123).
Why A is wrong: A is wrong because the force per unit length depends on the product I₁I₂, which is the same for both wires; an unequal current does not give an unequal force.
Why B is wrong: B is wrong for the same reason, and additionally contradicts Newton's third law, which holds for this interaction regardless of the current values.
Why D is wrong: D is wrong by a factor of ten in the arithmetic: (2 × 10⁻⁷ × 10.0)/0.200 = 1.00 × 10⁻⁵, not 1.00 × 10⁻⁴.
MCQ 6Concept TrapPractice
A stem reads: "Two long parallel wires 5.0 cm apart carry 3.0 A and 4.0 A in the same direction. Find the force on a 2.0 m length of one wire." A student computes μ₀I₁I₂/(2πd) and reports that value directly as the answer in newtons. The error is that
Show answer and why every option is right or wrong
Answer: B. μ₀I₁I₂/(2πd) is F/L in N/m; a stem that supplies a length is asking for the total force, so the result must be multiplied by that length (NCERT Class 12 Physics, Chapter 4, page 123).
Why A is wrong: A is wrong because the expression holds for any pair of steady currents; the symmetric product I₁I₂ handles unequal currents without modification.
Why C is wrong: C is wrong because the two wires feel equal and opposite forces — the expression already gives the force on each, and doubling it double-counts an action-reaction pair.
Why D is wrong: D is wrong because unit conversion is genuinely required (5.0 cm = 0.050 m), so the statement inverts the actual requirement; in any case that is not the student's stated error.
MCQ 7CalculationPractice
Two long parallel wires carry equal steady currents I and experience a force per unit length F₀. Both currents are now tripled and the separation is simultaneously tripled. The new force per unit length is
Show answer and why every option is right or wrong
Answer: B. F/L ∝ I₁I₂/d, so the numerator scales by 3 × 3 = 9 while the denominator scales by 3, giving 9/3 = 3 times the original, i.e. 3F₀ (NCERT Class 12 Physics, Chapter 4, page 123).
Why A is wrong: A is wrong because it assumes the current and separation factors cancel one-for-one; the currents contribute two factors of 3 and the separation only one.
Why C is wrong: C is wrong because it applies the 9× increase from the currents while ignoring the 3× increase in separation that partly offsets it.
Why D is wrong: D is wrong because it tracks only the separation's 1/d dependence and omits the current increase entirely.
MCQ 8CalculationPractice
Three long straight parallel wires P, Q and R lie in a plane in that order, with equal spacing d between neighbours. Each carries the same steady current I in the same direction. The net force per unit length on the middle wire Q is
Show answer and why every option is right or wrong
Answer: D. Q is attracted towards P and towards R with equal magnitudes μ₀I²/(2πd), since both currents and both separations are equal; the two attractions are oppositely directed and cancel (NCERT Class 12 Physics, Chapter 4, page 123).
Why A is wrong: A is wrong because it counts only the pull from P and ignores the equal and opposite pull from R on the other side.
Why B is wrong: B is wrong on both counts: the two pulls cancel rather than leaving a residue, and μ₀I²/(4πd) is in any case half the single-pair value.
Why C is wrong: C is wrong for the same reason as B, with the residual force additionally pointing the wrong way given the symmetry of the arrangement.
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Force Between Parallel Currents: quick recall before you leave
How do you solve a Force Between Parallel Currents question? A worked example
- 1
Given
Two long straight parallel wires in vacuum.
I₁ = 8.00 A, I₂ = 5.00 A, both in the same direction.
Separation d = 4.00 × 10⁻² m.
Length of wire under consideration L = 1.50 m.
μ₀ = 4π × 10⁻⁷ T·m/A (exact, by SI definition). - 2
Required
The magnitude of the force on the 1.50 m length of either wire, and whether the wires attract or repel.
- 3
Concept
Each wire sits in the magnetic field produced by the other and therefore experiences a force. For two long parallel conductors this reduces to a force per unit length that depends on the product of the currents and inversely on the separation. Currents in the same direction attract.
- 4
Formula
F/L = μ₀ I₁ I₂ / (2π d)
Total force on a length L: F = (μ₀ I₁ I₂ / (2π d)) × L - 5
Substitution
F/L = (4π × 10⁻⁷ × 8.00 × 5.00) / (2π × 4.00 × 10⁻²)
- 6
Calculation
The π cancels between numerator and denominator, and 4/2 = 2, leaving the standard reduced form:
F/L = (2 × 10⁻⁷ × 8.00 × 5.00) / (4.00 × 10⁻²)
F/L = (2 × 10⁻⁷ × 40.0) / (4.00 × 10⁻²)
F/L = (8.00 × 10⁻⁶) / (4.00 × 10⁻²) = 2.00 × 10⁻⁴ N/m
F = 2.00 × 10⁻⁴ × 1.50 = 3.00 × 10⁻⁴ N
μ₀ = 4π × 10⁻⁷ T·m/A is an exact defined constant and π is a mathematical constant, so neither contributes to the significant-figure count. The data carry three significant figures (8.00, 5.00, 4.00 × 10⁻², 1.50), so the answer is quoted to three. - 7
Final answer
F = 3.00 × 10⁻⁴ N, and the wires attract, since the currents are in the same direction.
- 8
Common trap
Two failures dominate here. The first is reporting 2.00 × 10⁻⁴ as the final answer in newtons — that is the force per metre, and the stem asked about a specific 1.50 m length. The second is writing "repel" out of the electrostatic reflex that like quantities repel; for currents, like directions attract.
- 9
Similar NEET-style question
Two long straight parallel wires in vacuum are separated by 2.50 × 10⁻² m and carry currents of 3.00 A and 12.0 A in opposite directions. Find the magnitude of the force on a 0.800 m length of one wire, and state whether the wires attract or repel. (Answer: 2.30 × 10⁻⁴ N; they repel.)
What to remember before solving Force Between Parallel Currents questions
Which Force Between Parallel Currents formulas do you need for NEET?
1 formula — click to collapse
Force between parallel currents
Force per unit length between long parallel wires separated by d. Same direction → attractive.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I1, I2 | currents | A |
| d | separation | m |
Valid when
- Long parallel wires
- Steady currents
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Force Between Parallel Currents questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
1μ₀I²/(2πλg)
2μ₀I²/(πλg)
32μ₀I²/(πλg)
44μ₀I²/(πλg)
NTA Answer: Option 2(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
Sources
NCERT refs: Class 12 Physics Chapter 4, p.123
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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